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lubasha [3.4K]
2 years ago
5

The equation f = v + at represents the final velocity of an object, f, with an initial velocity, v, and an acceleration rate, a,

over time, t.
Which is an equivalent equation solved for t?

t = t equals StartFraction f minus v Over a EndFraction z.
t = t equals StartFraction f minus a Over v EndFraction z.
t = a(f – v)
t = v(f – a)
Mathematics
1 answer:
Pavlova-9 [17]2 years ago
4 0

Answer:

t = t equals StartFraction f minus v Over a EndFraction z.

Step-by-step explanation:

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two bags contains 10 balls each. Every round you choose with equal probability one of the two bags and pick out a ball. After a
Stells [14]

Answer:

6.10%

Step-by-step explanation:

For every round since we are choosing with equal probability, then the probability of choosing from either of the bags is 1/2

Here, this scenario is only possible when a particular bag has been chosen 10 times and the other bag has been chosen 6 times ( meaning a particular bag has been emptied and for a particular bag to be emptied, all of its content would have been picked)

Now on the 17th trial, an empty bag is chosen

Therefore the required probability will be;

16C0 * (1/2)^10 * (1/2)^6 * 1/2 = 0.0610 = 6.1%

4 0
3 years ago
T(a+b)=r solve for a I need help due tomorrow
tigry1 [53]
Divide both sides by <span>t
</span>
<span>a+b=r/t 

</span>Subtract b <span>from both sides 
</span>
a<span>=r/t-b = Solution </span>
6 0
3 years ago
2/9 of the students in a school are in 6th grade.
Cerrena [4.2K]

Answer:

depended is the 6th grade

independed the 2/9 students

-by-step explanation:

4 0
2 years ago
A particle moves on a circle through points which been marked 0,1,2,3,4 (in a clockwise order). At each step it has a probabilit
Sedaia [141]

Answer:

Step-by-step explanation:

Given data:

SS={0,1,2,3,4}

Let probability of moving to the right be = P

Then probability of moving to the left is =1-P

The transition probability matrix is:

\left[\begin{array}{ccccc}1&P&0&0&0\\1-P&1&P&0&0\\0&1-P&1&P&0\\0&0&1-P&1&P\\0&0&0&1-P&1\end{array}\right]

Calculating the limiting probabilities:

π0=π0+Pπ1                 eq(1)

π1=(1-P)π0+π1+Pπ2     eq(2)

π2=(1-P)π1+π2+Pπ3    eq(3)

π3=(1-P)π2+π3+Pπ4    eq(4)

π4=(1-P)π3+π4             eq(5)

π0+π1+π2+π3+π4=1

π0-π0-Pπ1=0

→π1 = 0

substituting value of π1  in eq(2)

(1-P)π0+Pπ2=0

from

π2=(1-P)π1+π2+Pπ3  

we get

(1-P)π1+Pπ3 = 0

from

π3=(1-P)π2+π3+Pπ4

we get

(1-P)π2+Pπ4 =0

from π4=(1-P)π3+π4  

→π3=0

substituting values of π1 and π3 in eq(3)

→π2=0

Now

π0+π1+π2+π3+π4=0

π0+π4=1

π0=0.5

π4=0.5

So limiting probabilities are {0.5,0,0,0,0.5}

4 0
3 years ago
Graph the liner equation for y=x÷3
KonstantinChe [14]
' x ÷ 3 ' is just another way to write  ' 1/3 x ' ,

so your equation is        y = 1/3 x .

The graph of that equation is a straight line,
going through the origin, with a slope of 1/3 .

3 0
2 years ago
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